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3 years ago

_16. Which expression is equivalent to

Mathematics

1 answer:

Phantasy [73]3 years ago

5 0

Answer:

-2x-8

Step-by-step explanation:

1) Distribute:

-6/3x+-15/3-3

2) Simplify:

-2x-5-3

3) combine like terms:

-2x-8

hope this helps

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Determine the sample size needed to construct a 99% confidence interval to estimate the average GPA for the student population

Volgvan

Answer: n = 14

Step-by-step explanation: margin of error = critical value × σ/√n

Where σ = population standard deviation = 1

n = sample size = ?

We are to construct a 99% confidence interval, hence the level of significance is 1%.

The critical value for 2 tailed test at 1% level of significance is gotten from a standard normal distribution table which is 2.58

Margin of error = 0.7

0.7 = 2.58×1/√n

0.7 = 2.58/√n

By cross multipying

0.7×√n = 2.58

By squaring both sides

0.7^2 × n = 2.58^2

0.49 × n = 6.6564

n = 6.6564/0.49

n = 14

3 0

3 years ago

Solve the equation 4/x=8

Veseljchak [2.6K]

Answer:

x = .5

Step-by-step explanation:

4 / x = 8

4 / .5 = 8

x = .5

7 0

2 years ago

Square sandbox with sides 3 feet long. She wants to put sand 0.85 feet deep in the box. What is the exact measure, in cubic feet

Blababa [14]

3 * 3 = 9 * 0.85 = 7.65 cubic feet of sand

6 0

3 years ago

The amount of protein that an individual must consume is different for every person. There are solid theoretical ideas that sugg

amid [387]

Answer:

The proportion of the population that have a protein requirement less than 0.60 g P • kg-1 • d-1 is 0.239, that is, 239 persons for every 1000, or simply 23.9% of them.

$\\ 0.239 =\frac{239}{1000}\;or\;23.9\%$

Step-by-step explanation:

From the question, we have the following information:

The distribution for protein requirement is normally distributed.
The population mean for protein requirement for adults is $\\ \mu= 0.65 gP*kg^{-1}*d^{-1}$
The population standard deviation is $\\ \sigma =0.07 gP*kg^{-1}*d^{-1}$

We have here that protein requirements in adults is normally distributed with defined parameters. The question is about the proportion of the population that has a requirement less than $\\ x = 0.60 gP*kg^{-1}*d^{-1}$ .

For answering this, we need to calculate a z-score to obtain the probability of the value x in this distribution using a standard normal table available on the Internet or on any statistics book.

<h3>z-score</h3>

A z-score is expressed as

$\\ z = \frac{x - \mu}{\sigma}$

For the given parameters, we have:

$\\ z = \frac{0.60 - 0.65}{0.07}$

$\\ z = -0.7142857$

<h3>Determining the probability</h3>

With this value for z at hand, we need to consult a standard normal table to determine what the probability of this value is.

The value for z = -0.7142857 is telling us that the requirement for protein is below the population mean (negative sign indicates this). However, most standard normal tables give a probability that a statistic is less than z and for values greater than the mean (in other words, positive values). To overcome this, we need to take the complement of the probability given for z-score z = 0.7142857, that is, subtract from 1 this probability, which is possible because the normal distribution is symmetrical.

Tables have values for z with two decimal places, then, for z = 0.7142857, we need to rewrite it as z = 0.71. For this value, the standard normal table gives a value of P(z<0.71) = 0.76115.

Therefore, the cumulative probability for values less than x = 0.60 which corresponds to a z-score = -0.7142857 is approximately:

$\\ P(x$